Solve x/12+3lessthanequalto7

Stella [2.4K]

Answer:

The smaller one is x = 3
and the larger one is x = 6

Step-by-step explanation:

A "critical number" is a function argument where the function has zero or undefined slope. A polynomial function never has undefined slope, so we're looking for the x-values where f'(x) = 0.

A graphing calculator easily shows the points that have zero slope. They are found at x = 3 and x = 6.

_____

Setting the derivative to zero is another way to find the points with zero slope:

f'(x) = 6x^2 -54x +108 = 6(x^2 -9x +18) = 6(x-3)(x-6)

The derivative will be zero where the factors are zero, at x=3 and x=6.

The critical numbers are x=3 and x=6.

6 0

2 years ago

Chapter 7, Section 7.2, Question P9 The process for manufacturing a ball bearing results in weights that have an approximately n

bogdanovich [222]

a. Let $X$ be a random variable representing the weight of a ball bearing selected at random. We're told that $X\sim\mathcal N(0.15,0.005^2)$ , so

$\mathrm P(X$

where $Z\sim\mathcal N(0,1)$ . This probability is approximately

$\mathrm P(Z$

b. Let $X_i$ be a random variable representing the weight of the $i$ -th ball that is selected, and let $Y$ be the mean of these 4 weights,

$Y=\dfrac{X_1+X_2+X_3+X_4}4$

The sum of normally distributed random variables is a random variable that also follows a normal distribution,

$X_1+X_2+X_3+X_4\sim\mathcal N(4\cdot0.15,4\cdot0.005^2)$

so that

$Y\sim\mathcal N(0.15,0.005^2)$

Then

$\mathrm P(Y$

c. Same as (b).

8 0

3 years ago

A family of 3 children 2 adults visited a heath club. The club charges $y an hour for each child and $x an hour for each adult.

gregori [183]

Answer:

Step-by-step explanation:y<30 and x<15

6 0

3 years ago

A rancher wishes to build a fence to enclose a 2250 square yard rectangular field. Along one side the fence is to be made of hea

Bess [88]

Answer:

The least cost of fencing for the rancher is $1200

Step-by-step explanation:

Let x be the width and y the length of the rectangular field.

Let C the total cost of the rectangular field.

The side made of heavy duty material of length of x costs 16 dollars a yard. The three sides not made of heavy duty material cost $4 per yard, their side lengths are x, y, y. Thus

$C=4x+4y+4y+16x\\C=20x+8y$

We know that the total area of rectangular field should be 2250 square yards,

$x\cdot y=2250$

We can say that $y=\frac{2250}{x}$

Substituting into the total cost of the rectangular field, we get

$C=20x+8(\frac{2250}{x})\\\\C=20x+\frac{18000}{x}$

We have to figure out where the function is increasing and decreasing. Differentiating,

$\frac{d}{dx}C=\frac{d}{dx}\left(20x+\frac{18000}{x}\right)\\\\C'=20-\frac{18000}{x^2}$

Next, we find the critical points of the derivative

$20-\frac{18000}{x^2}=0\\\\20x^2-\frac{18000}{x^2}x^2=0\cdot \:x^2\\\\20x^2-18000=0\\\\20x^2-18000+18000=0+18000\\\\20x^2=18000\\\\\frac{20x^2}{20}=\frac{18000}{20}\\\\x^2=900\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{900},\:x=-\sqrt{900}\\\\x=30,\:x=-30$

Because the length is always positive the only point we take is $x=30$ . We thus test the intervals $(0, 30)$ and $(30, \infty)$

$C'(20)=20-\frac{18000}{20^2} = -25 < 0\\\\C'(40)= 20-\frac{18000}{20^2} = 8.75 >0$

we see that total cost function is decreasing on $(0, 30)$ and increasing on $(30, \infty)$ . Therefore, the minimum is attained at $x=30$ , so the minimal cost is

$C(30)=20(30)+\frac{18000}{30}\\C(30)=1200$

The least cost of fencing for the rancher is $1200

Here’s the diagram:

3 0

2 years ago

A right triangle has legs measuring 4.5 meters and 1.5 meters. The lenghts of the legs of a second triangle are proportional to

Eduardwww [97]

We know that
in the first triangle
the ratio of the legs are
4.5/1.5-----> 3

then
case A) 6 m and 2 m ------> ratio=6/3----> 3
so
the legs of a second triangle are proportional to the lengths of the legs of the first triangle

case B) 8 m and 5 m ------> ratio=8/5---->1.6
so
the legs of a second triangle are not proportional to the lengths of the legs of the first triangle

case C) 7 m and 3.5 mm ------> ratio=7/3.5---->2
so
the legs of a second triangle are not proportional to the lengths of the legs of the first triangle

case D) 10 m and 2.5 m ------> ratio=10/2.5---->4
so
the legs of a second triangle are not proportional to the lengths of the legs of the first triangle

case E) 11.25 m and 3.75 m ------> ratio=11.25/3.75---->3
so
the legs of a second triangle are proportional to the lengths of the legs of the first triangle

the answer is
A) 6 m and 2 m
E) 11.25 m and 3.75 m

4 0

3 years ago